calculate e.m.f of a cell made up of a standard O2 half-cell
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calculate e.m.f of a cell made up of a standard O2 half-cell in a basic solution (PH=14) and a half-cell obtained by dipping at Pt electrode into a 0.46 M Fe 2+ ion and 1.0M Fe 3+ ion solution ( electrode of the second kind)
楼下英文很好,我是翻译不出来,外文:1M =1mol/L
计算电动势:在pH=14的碱性半电池中通氧气,另一半电池中将铂电极浸入含有0.46mol/L 亚铁离子和1.0mol/L 三价铁离子的溶液
负极:O2 +4e- + 2H2O=4OH- φ0 (O2/OH-)=0.401v
正极:Fe3+ + e-=Fe2+ φ0 (Fe3+/Fe2+)=0.771v
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由于氧气的气压不知,就当它是标压p0=100kPa,c(OH-)=1mol/L (pH=14)
φ(O2/OH-)= φ0 (O2/OH-)=0.401v
又:φ(Fe3+/Fe2+)=φ0 (Fe3+/Fe2+)+0.0591v ×lg c(Fe3+)/c(Fe2+)
=φ0 (Fe3+/Fe2+)+0.0591v ×lg 1.0/ 0.46
=0.791v
因此电动势:E=φ(Fe3+/Fe2+)-φ(O2/OH-)
=0.791v -0.401v
=0.39v
计算电动势:在pH=14的碱性半电池中通氧气,另一半电池中将铂电极浸入含有0.46mol/L 亚铁离子和1.0mol/L 三价铁离子的溶液
负极:O2 +4e- + 2H2O=4OH- φ0 (O2/OH-)=0.401v
正极:Fe3+ + e-=Fe2+ φ0 (Fe3+/Fe2+)=0.771v
---
由于氧气的气压不知,就当它是标压p0=100kPa,c(OH-)=1mol/L (pH=14)
φ(O2/OH-)= φ0 (O2/OH-)=0.401v
又:φ(Fe3+/Fe2+)=φ0 (Fe3+/Fe2+)+0.0591v ×lg c(Fe3+)/c(Fe2+)
=φ0 (Fe3+/Fe2+)+0.0591v ×lg 1.0/ 0.46
=0.791v
因此电动势:E=φ(Fe3+/Fe2+)-φ(O2/OH-)
=0.791v -0.401v
=0.39v
calculate e.m.f of a cell made up of a standard O2 half-cell
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